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$$\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$$

  1. First step would be proving that the statement is true for n=2

On the LHS for $n=2$ we would have $\frac{1}{4}$ therefore the statement is true for $n=2$

  1. Now we must assume the statement is true for $n=j$ with $j\geq2$

$$\sum\limits_{k=2}^{j}{\frac{1}{k^2}}<1$$

  1. Now we must prove true for $n=j+1$

$\sum\limits_{k=2}^{j+1}{\frac{1}{k^2}}<1$

$\Rightarrow \sum\limits_{k=2}^{j}{\frac{1}{k^2}}+\frac{1}{(k+1)^2}<1$

I cant seem to introduce the induction hypothesis

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    $\begingroup$ "Blindly" using induction would clearly not work since the RHS doesn't increase. So what you want to do is "stronger" induction where you strengthen the hypothesis. $\endgroup$
    – Calvin Lin
    Apr 26, 2020 at 16:46

4 Answers 4

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Let $ n\in\mathbb{N} : $

Notice that for any integer $ k\geq 2 $, we have $ k^{2}\geq k\left(k-1\right) $, and thus : $$ \sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\sum_{k=2}^{n}{\frac{1}{k\left(k-1\right)}}=\sum_{k=2}^{n}{\left(\frac{1}{k-1}-\frac{1}{k}\right)}=1-\frac{1}{n}<1 $$

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    $\begingroup$ (+1) for the "classical" approach $\endgroup$
    – Mark Viola
    Apr 26, 2020 at 17:03
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Strengthen the induction hypothesis

Hint: Prove by induction that

$$ \sum_{k=2}^n \frac{1}{k^2 } < 1 - \frac{1}{n}.$$

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  • $\begingroup$ Im not sure I quite follow. How is this equivalent to the initial statement? $\endgroup$ Apr 26, 2020 at 16:54
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    $\begingroup$ They are not equivalent. However, if you prove the inequality of @Calvin Lin, then, you are done. As an analogy, suppose you need to show that $x \leq 3$. Certainly, one way to show this is that $x \leq 2$:) It is the same logic here. Also just as a side note if you use the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$. Then, the result follows, as well. Of course, it is an overkill for this question. Since we are using a fact which is more difficult than the question itself. $\endgroup$
    – ALNS
    Apr 26, 2020 at 16:59
  • $\begingroup$ @MarkViola Thanks, fixed the typo. IMO For "obvious" typos like this, you can go ahead and edit the writeup. $\endgroup$
    – Calvin Lin
    Apr 26, 2020 at 17:04
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Here is another approach :

We can easily prove that $ \left(\forall x\in\left[0,\frac{1}{2}\right]\right),\ \frac{2}{\pi}\arctan{\left(2x\right)}-x\geq 0 $ (By differentiation the function $ x\mapsto\frac{2}{\pi}\arctan{\left(2x\right)}-x $, studying its variations, etc...)

Thus, if $ n $ is a positive integer, we have the following : \begin{aligned}\sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{2}{k^{2}}\right)}}&=\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{\left(k+1\right)-\left(k-1\right)}{1+\left(k+1\right)\left(k-1\right)}\right)}}\\&=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{k}\right)}+\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{k}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\left(\arctan{\left(n+1\right)}-\arctan{2}\right)+\frac{2}{\pi}\left(\arctan{n}-\frac{\pi}{4}\right)\\ &\leq\frac{3}{2}-\frac{2}{\pi}\arctan{2}<1\end{aligned}

To get to the last line we upper bounded the $ \arctan $s by $ \frac{\pi}{2} \cdot $

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$f(x)= \frac{1}{x^2}$, $x >0$, stricly decreasing.

$ \displaystyle{\sum_{k=2}^{n}}\frac{1}{k^2} <\displaystyle{ \int_{1}^{n}}\frac{1}{x^2}dx =-\frac{1}{x}\big ]_1^n=$

$1-1/n. $

As suggested by Calvin you can use the above inequality for the induction proof.

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  • $\begingroup$ This was already posted 5 minutes ago $\endgroup$
    – Mark Viola
    Apr 26, 2020 at 17:04

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